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Chapter 4: Problem 22

A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

Short Answer

Expert verified
(a) 0.494 seconds; (b) 3.08 m/s.

Step by step solution

01

Determine the time in the air using vertical motion

Use the equation for vertical motion, assuming initial vertical velocity is zero. The equation is \[ h = \frac{1}{2} g t^2 \]where \( h = 1.20 \mathrm{~m} \) is the height, and \( g = 9.81 \mathrm{~m/s^2} \) is the acceleration due to gravity. Solve for \( t \). \[ t^2 = \frac{2h}{g} = \frac{2 \times 1.20}{9.81} \approx 0.244 \] \[ t = \sqrt{0.244} \approx 0.494 \text{ seconds} \] Thus, the ball is in the air for approximately 0.494 seconds.
02

Calculate the horizontal speed using horizontal motion

Now calculate the horizontal speed using the formula for horizontal motion: \[ d = v_{x} t \] where \( d = 1.52 \mathrm{~m} \) is the horizontal distance and \( t = 0.494 \text{ sec} \) is the time calculated before. Solve for \( v_{x} \): \[ v_{x} = \frac{d}{t} = \frac{1.52}{0.494} \approx 3.08 \mathrm{~m/s} \] So, the ball's speed at the instant it leaves the table is approximately 3.08 m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Motion
Understanding vertical motion is key when dealing with projectile problems like this one. When an object rolls off a table, it starts moving downward due to gravity. This motion is vertical because gravity acts directly downwards, accelerating the object toward the ground. In our exercise, the height from which the ball falls is given as 1.20 meters. The formula to determine how long the object stays in the air before hitting the ground is\[ h = \frac{1}{2} g t^2 \]Here, \( t \) is the time in seconds, \( g \) is the acceleration due to gravity (9.81 m/s虏), and \( h \) is the height from which the object falls. Since the ball begins its descent from rest, the initial vertical speed is 0.
Horizontal Motion
Horizontal motion in projectile problems is independent of vertical motion. This means that while the ball falls, it also moves horizontally at a constant speed, because no external forces are acting in that direction (ignoring air resistance for simplicity). The horizontal distance traveled is calculated using the formula:\[ d = v_{x} t \]Where \( d \) is the horizontal distance (1.52 meters, in this case), \( v_{x} \) is the horizontal speed of the ball as it leaves the edge of the table, and \( t \) is the time calculated from vertical motion (0.494 seconds). The formula reveals the direct dependence between horizontal speed, time, and distance.
Acceleration Due to Gravity
Gravity is the force that pulls objects toward the Earth, and it impacts the vertical motion of the ball in our exercise. The standard gravitational acceleration is approximately 9.81 m/s虏 on Earth, and it affects only the vertical component of the ball's motion. It consistently accelerates the ball downward once it starts rolling off the edge of the table. 鈥� Gravity acts independently of horizontal motion. 鈥� Acceleration due to gravity is constant at 9.81 m/s虏. 鈥� It determines how fast an object will accelerate towards the ground from rest.

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Most popular questions from this chapter

A ball rolls horizontally off the top of a stairway with a speed of \(1.52 \mathrm{~m} / \mathrm{s}\). The steps are \(20.3 \mathrm{~cm}\) high and \(20.3 \mathrm{~cm}\) wide. Which step does the ball hit first?

A lowly high diver pushes off horizontally with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) from the platform edge \(10.0 \mathrm{~m}\) above the surface of the water. (a) At what horizontal distance from the edge is the diver \(0.800 \mathrm{~s}\) after pushing off? (b) At what vertical distance above the surface of the water is the diver just then? (c) At what horizontal distance from the edge does the diver strike the water?

A proton initially has \(\vec{v}=4.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and then \(4.0 \mathrm{~s}\) later has \(\vec{v}=-2.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+5.0 \hat{\mathrm{k}}\) (in meters per second). For that \(4.0 \mathrm{~s}\), what are (a) the proton's average acceleration \(\vec{a}_{\text {avg }}\) in unitvector notation, (b) the magnitude of \(\vec{a}_{\text {avg }}\), and \((\) c) the angle between \(\vec{a}_{\text {avg }}\) and the positive direction of the \(x\) axis?

A wooden boxcar is moving along a stra?ght ralroad track at speed \(v_{1}\). A sniper fires a bullet (initial speed \(v_{2}\) ) at it from a high-powered rifle. The bullet passes through both lengthwise walls of the car, its entrance and exit holes being exactly opposite each other as viewed from within the car. From what direction, relative to the track, is the bullet fired? Assume that the bullet is not deflected upon entering the car, but that its speed decreases by \(20 \%\). Take \(v_{1}=85 \mathrm{~km} / \mathrm{h}\) and \(v_{2}=650 \mathrm{~m} / \mathrm{s}\). (Why don't you need to know the width of the boxcar?)

A dart is thrown horizontally with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) toward point \(P\), the bull's-eye on a dart board. It hits at point \(Q\) on the rim, vertically below \(P, 0.19 \mathrm{~s}\) later. (a) What is the distance \(P Q ?\) (b) How far away from the dart board is the dart released?
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